Pass PHP array through Select Option Fields

Question

I am writing a basic CMS system and have come across something which should be seemingly simple -but is beginning to frustrate me.!

I am trying to pass an array through a select option field to populate a list of categories in which I can save a post.

I have a 'posts' form which comprises of 3 fields. Title, content and Category ID (CatID).

When the user creates a post, they can select the category they wish to assign the post assigned to by using a drop down list - (this is populated by using a different form).

So the technical bit; -

MySQL DB:-

categories = catname (char60 PRIMARY), catid (INT10, AI) posts = id (bigint20 PRIMARY), catid (int10 PRIMARY), title (text), content (varchar255)

Example of categories populates: catname = Home / catid = 1 ...etc

Output.php ;

<?php
function display_post_form($post = '') {
$edit = is_array($post);
?>
    <form action="<?php echo $edit ? 'edit.php' : 'add.php' ; ?>" method="post">
    <table border="0">
    <tr>
    <td> Title:</td>
        <td> <input type="text" name="title" value="<?php echo $edit ? $post['title'] : '' ; ?>" size="60" /> </td>
    </tr><tr>
    <td> Content:</td>
    <td> <textarea id="editor1" name="content" value="<?php echo $edit ? $post['content'] : '' ; ?>"> </textarea> </td>
    </tr><tr>
    <td> Category:</td>
    <td><select name="catid">
    <?php
            $cat_array = get_categories($catid, $catname);
        foreach($cat_array as $thiscat) {
        echo "<option value=\"".$thiscat['catid']."\" ";
            if (($edit) && ($thiscat['catid'] == $post['catid'])) {
            echo " selected";
        }
        echo ">".$thiscat['catname']."</option>";
        }
    ?>
    </select>
    </td>
    </tr><tr>
    <td> Button:</td>
    <td <?php if (!$edit) { echo "colspan=2"; } ?> align="center">
    <?php
            if ($edit)
        echo "<input type=\"hidden\" name=\"_id\" value=\"". $post['id'] ."\" />";
    ?>
    <input type="submit" value="<?php echo $edit ? 'Update' : 'Add' ; ?> Post" />
    </form></td>
        </td>
        </tr>
    </table>
</form>
<?php
}
?>

Functions.php ;

function get_categories($catid, $catname) {

$conn = db_connect();

if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL "  .mysqli_connect_error();
}

$sql = "SELECT catname, catid FROM categories";

$result = mysqli_query($conn, $sql) or die(" Could not query database");

    while($row = mysqli_fetch_assoc($result)) {
    printf("\n %s %s |\n",$row["catname"],$row["catid"]);
}

mysqli_close($conn);

}

I am able to call in the 'get_cattegories()' function which generates a flat data of categories and their respective id's. I then combined this with the Select Option Field in the Output.php file and it doesn't generate anything.

Can anyone give some useful tips or advice? Many thanks :)

Why are you putting parameters in get_gategories function? Where are you returning anything from that function? Why don't you just put catname and catid in an array and return it? — AR.
– AR., Commented Mar 20, 2014 at 0:06

Filip Górny · Accepted Answer · 2014-03-21 07:42:04Z

2

You are not returning the array but printing a string to the output. Change printf to return:

function get_categories($catid, $catname) {

$conn = db_connect();

if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL "  .mysqli_connect_error();
}

$sql = "SELECT catname, catid FROM categories";

$result = mysqli_query($conn, $sql) or die(" Could not query database");
$categories = array();
    while($row = mysqli_fetch_assoc($result)) {
    $categories[] = $row;
}

mysqli_close($conn);

return $categories;

}

Also I agree for the comments to your question. The arguments are useless.

You also may refactor the code, actually... alot. Move the mysql_connect() to the other place, probably at the beginning of your script.

I suggest to use some frameworks. I think KohanaPHP will be a good start. You will learn about architecture and some design patterns. Keep the good work and improve your skills ;-)

edited Mar 21, 2014 at 7:42

answered Mar 20, 2014 at 0:07

Filip Górny

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